How do you integrate $\int x^{3} t a n x d x$ $int x^3 t an x dx$ using integration by parts?

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How do you integrate $\int x^{3} t a n x d x$ $int x^3 t an x dx$ using integration by parts?

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Answer 1 · 2016-01-07T01:50:05

$\int x^{3} tan (x) d x = ln | cos (x) | (x^{3} - 3 x^{2} + 6 x - 6 + c)$ $intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)$

Explanation:

Say $d v = tan (x)$ $dv = tan(x)$ so $v = ln | cos (x) |$ $v = ln|cos(x)|$
And $u = x^{3}$ $u = x^3$ so $d u = 3 x^{2}$ $du = 3x^2$

$\int x^{3} tan (x) d x = x^{3} \cdot ln | cos (x) | - 3 \int tan (x) \cdot x^{2} d x$ $intx^3tan(x)dx = x^3*ln|cos(x)| - 3inttan(x)*x^2dx$

For the latter integral, repeat

$d v = tan (x)$ $dv = tan(x)$ so $v = ln | cos (x) |$ $v = ln|cos(x)|$ , $u = x^{2}$ $u = x^2$ so $d u = 2 x$ $du = 2x$

$\int x^{3} tan (x) d x = x^{3} \cdot ln | cos (x) | - 3 (x^{2} ln | cos (x) | - 2 \int tan (x) x d x)$ $intx^3tan(x)dx = x^3*ln|cos(x)| - 3(x^2ln|cos(x)| - 2inttan(x)xdx)$
$\int x^{3} tan (x) d x = x^{3} \cdot ln | cos (x) | - 3 x^{2} ln | cos (x) | + 6 \int tan (x) x d x$ $intx^3tan(x)dx = x^3*ln|cos(x)| - 3x^2ln|cos(x)| + 6inttan(x)xdx$

And once more to finish it

$d v = tan (x)$ $dv = tan(x)$ so $v = ln | cos (x) |$ $v = ln|cos(x)|$ , $u = x$ $u = x$ so $d u = 1$ $du = 1$

$\int x^{3} tan (x) d x = x^{3} ln | cos (x) | - 3 x^{2} ln | cos (x) | + 6 (x ln | cos (x) | - ln | cos (x) | + c)$ $intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6(xln|cos(x)| - ln|cos(x)| + c)$

$\int x^{3} tan (x) d x = x^{3} ln | cos (x) | - 3 x^{2} ln | cos (x) | + 6 x ln | cos (x) | - 6 ln | cos (x) | + c$ $intx^3tan(x)dx = x^3ln|cos(x)| - 3x^2ln|cos(x)| + 6xln|cos(x)| - 6ln|cos(x)| + c$

Or

$\int x^{3} tan (x) d x = ln | cos (x) | (x^{3} - 3 x^{2} + 6 x - 6 + c)$ $intx^3tan(x)dx = ln|cos(x)|(x^3 - 3x^2 + 6x - 6 + c)$

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How do you integrate $\int x^{3} t a n x d x$ $int x^3 t an x dx$ using integration by parts?

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